Results 1 to 4 of 4

Math Help - Eigenvalues of Linear operator

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    184

    Eigenvalues of Linear operator

    Hello,
    Find all the eigenvalues and determine with a reason whether L is self-adjoint.
    " alt="d/dt" />----------> L_2[0,1] where the domain D is given by D{f belongs to L_2[0,1]: f is continuously differentiable and f(0)=0)}
    I get f(x)=e^{\lambda x} after solving first order linear differential equation. How do i find \lambda?
    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,444
    Thanks
    1863
    Quote Originally Posted by charikaar View Post
    Hello,
    Find all the eigenvalues and determine with a reason whether L is self-adjoint.
    " alt="d/dt" />----------> L_2[0,1] where the domain D is given by D{f belongs to L_2[0,1]: f is continuously differentiable and f(0)=0)}
    I get f(x)=e^{\lambda x} after solving first order linear differential equation. How do i find \lambda?
    thanks
    our operator is just differentiation so, yes, \frac{d e^{\lambda x}}{dx}= \lambda e^{\lambda x}. More generally, if \frac{dy}{dx}= \lambda y, then y= Ce^{\lambda 0} e^{\lambda x} is an "eigenvector" and \lambda is an eigenvalue. As for finding \lambda, you don't! That equation is true for all \lambda- any real number is an eigenvalue.

    As for whether or not it is "self adjoint", the definition of self-adjoint is that <Af, g>= <f, Ag> for any vectors f, g. Here, I presume you are using the "standard" inner product, <f, g>= \int_0^1 f(x)g(x)dx so that we want \int_0^x \frac{df}{dx} g dt. Use "integration by parts" with u= g and dv= (df/dx)dx. Then du= (dg/dx)dx and v= f.

    \int u dv= uv- \int vdu= f(x)g(x)- \int f (dg/dx) dx= <f, dg/dx> provided f(1)g(1)- f(0)g(0)= 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    I'm curious on one thing. Don't we require that f(0) = 0?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2008
    Posts
    184
    Thanks for your help.

    L_2[0,1] is the Hilbert space of complex square integrable functions on [0,1] and D={f belongs to L_{2}[0,1] : f is continuously differentiable and f(0)=1)}. so \lambda includes complex numbers also? L is not self-adjoint because there is minus in front of the integral right?

    f(0)=1 not zero.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Eigenvalues of a linear operator
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: January 13th 2011, 12:13 PM
  2. Eigenvalues of Linear operator
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 12th 2010, 01:38 PM
  3. Real eigenvalues of linear operator
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 30th 2010, 02:48 PM
  4. linear operator eigenvalues
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 26th 2010, 04:54 PM
  5. Composition linear operator eigenvalues
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 20th 2009, 12:37 PM

Search Tags


/mathhelpforum @mathhelpforum